• 贪心，按照最晚下落时间点升序，下落时长降序排序，依次停放，对拍 wa了。

• 观察数据范围：N≤10，枚举所有排列，判断是否有合法的排列，复杂度 $O(n !*n * t)$，3e8的计算量大概1s，题目限制2s，稳！

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10, INF = 0x3f3f3f3f;
struct T {
    int t, d, l;
    bool operator<(const T& r) const {
        if (t + d != r.t + r.d)
            return t + d < r.t + r.d;
        return l < r.l;
    }
} t[N];

int main1() {
    int T, n, a, b, c;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d%d%d", &a, &b, &c);
            t[i] = {a, b, c};
        }
        sort(t + 1, t + 1 + n);
        int te = t[1].t + t[1].l, f = 1;
        for (int i = 2; i <= n; i++) {
            if (te > t[i].t + t[i].d) {
                f = 0;
                break;
            }
            if (te < t[i].t)
                te = t[i].t;
            te += t[i].l;
        }
        printf("%s\n", f ? "YES" : "NO");
    }
    return 0;
}

int p[N];
int main() {
    int T, n, a, b, c;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d%d%d", &a, &b, &c);
            t[i] = {a, b, c}, p[i] = i;
        }
        int te = 0, f = 1;
        do {
            te = 0, f = 1;
            for (int j = 1; j <= n; j++) {
                int i = p[j];
                if (te > t[i].t + t[i].d) {
                    f = 0;
                    break;
                }
                if (te < t[i].t)
                    te = t[i].t;
                te += t[i].l;
            }
            if (f)
                break;
        } while (next_permutation(p + 1, p + 1 + n));
        printf("%s\n", f ? "YES" : "NO");
    }
    return 0;
}